Answer:
a) i = -9.63 cm
, h ’= .0.24075 cm erect
b) i = 259.74 cm
,
Explanation:
For this exercise let's start by finding the focal length of the lens
1 / f = (n-1) (1 / R₁ - 1 / R₂)
1 / f = (1.70 -1)) 1 / ∞ - 1/13)
1 / f = 0.0538
f = - 18.57 cm
Now we can use the constructor equation
1 / f = 1 / o + 1 / i
1 / i = 1 / f - 1 / o
1 / i = -1 / 18.57 -1/20
1 / i = -0.1038 cm
I = -9.63 cm
For the height of the
image let's use magnification
m = h '/ h = - i / o
h ’= -h i / o
h ’= - 0.5 (-9.63) / 20
h ’= .0.24075 cm
b) we invert the lens
The focal length is
1 / f = (1.70 -1) (1/13 - 1 / int)
1 / f = 0.0538
f = 18.57 cm
1 / i = 1 / f -1 / o
1 / I = 1 / 18.57 - 1/20
1 / I = 3.85 10-3
i = 259.74 cm
h ’= - 0.5 259.74 / 20
h ’= 6.4935 cm
Answer:
56 kg
Explanation:
The change in potential energy of the man is given by:
where
m is the man's mass
g is the gravitational acceleration
is the change in height of the man
In this problem, we have:
is the gain in potential energy
g = 9.8 m/s^2 is the gravitational acceleration
is the change in height
Re-arranging the equation and substituting the numbers, we find the mass:
Answer:
The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During that time the voltage will drop from 9V to 6V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval?
Explanation:
